A function $f:\mathbb{R}^n\longrightarrow\mathbb{R}$ induces a measure by its finite differences, that is \begin{align} \mu((\mathbf{a},\mathbf{b}]) := \Delta_{a_1,b_1}\cdots\Delta_{a_n,b_n} f \end{align} where $\Delta_{a_1,b_1}f = f(b_1,a_2,...,a_n)f(a_1,a_1,...,a_n)$, if it is supermodular (positive finite differences of order $n$). Does anyone have a proof of this? I have seen this in some papers but without proof nor references.
Take the 2minute tour
×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.
